302^{x} + 2^{\left|x \right|}\geq 2\sqrt{2}
\text{When } x\geq 0,
2^{x+1}\geq 2^{3/2}
\Rightarrow x+1\geq \frac{3}{2} \Rightarrow x\geq \frac{1}{2}
\text{When } x<0,
2^{x}+2^{-x}\geq 2^{3/2}
\text{Take }2^{x}\text{ = t,}
t^{2}-2\sqrt{2}t +1\geq 0
\Rightarrow 8t^{2}-4\leq 0\Rightarrow t^{2}\leq 1/2
\Rightarrow \frac{-1}{\sqrt{2}}\leq t\leq \frac{1}{\sqrt{2}}
\text{But }2^{x}\geq 0
2^{x}\leq 2^{-1/2}\Rightarrow x\leq -1/2
\boxed{x\in(-\infty,-\frac{1}{2}]\cup[\frac{1}{2},\infty)}
262ashish u hv made a mistake in ur 2nd portion
when x<0
2x +2-x >=2√2
let 2x =t
t+1/t >= 2√2
or, t2 - 2√2t +1 >=0
or t>=√2 -1
or 2x >√2 -1
taking log, we have,
x<log2(√2 -1) ≈ 1.27
30
Ashish Kothari
·2011-06-17 03:00:06
You're correct, I messed up this one totally. [3]
t^{2}-2\sqrt{2}t+1\geq 0
\Rightarrow \left( t-\sqrt{2}\right)^{2}\geq 1
\Rightarrow \left(t-\sqrt{2} \right)\geq 1 \text{ or } \left(t-\sqrt{2} \right)\leq -1
\Rightarrow t\leq \sqrt{2}-1 \text{ or } t\geq \sqrt{2}+1
\Rightarrow 2^{x}\leq \sqrt{2}-1 \text{ or } 2^{x}\geq \sqrt{2}+1
\Rightarrow x\leq \log_{2} \left(\sqrt{2}- 1\right)
x \in \left( -\infty,\log_{2} \left(\sqrt{2}-1\right)\right] \cup \left[ \frac{1}{2},\infty\right)
I guess this should be the answer. That is if I'm not making some other stupid mistake again! :P
1yes ashish and aditya u r correct and thanx for explanation.......